show that ^n c_(r+1) +^n c_(r ) =^(n+1) c


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Question Number 66262 by peter frank last updated on 11/Aug/19

$s h o w t h a t$ $^{n} c_{r + 1} +^{n} c_{r} =^{n + 1} c_{r + 1}$
Commented by mathmax by abdo last updated on 12/Aug/19
$C_n ^r +C_n ^(r+1) =((n!)/(r!(n−r)!)) +((n!)/((r+1)!(n−r−1)!)) =((n!(r+1))/((r+1)!(n−r)!)) +((n!(n−r))/((r+1)!(n−r)!)) =((n!)/((r+1)!(n−r)!)){(r+1) +(n−r)} =(((n+1)!)/((r+1)!((n+1)−(r+1))!)) =C_(n+1) ^(r+1)$
$C_{n}^{r} + C_{n}^{r + 1} = \frac{n!}{r! (n - r)!} + \frac{n!}{(r + 1)! (n - r - 1)!}$ $= \frac{n! (r + 1)}{(r + 1)! (n - r)!} + \frac{n! (n - r)}{(r + 1)! (n - r)!}$ $= \frac{n!}{(r + 1)! (n - r)!} {(r + 1) + (n - r)} = \frac{(n + 1)!}{(r + 1)! ((n + 1) - (r + 1))!}$ $= C_{n + 1}^{r + 1}$
Commented by peter frank last updated on 12/Aug/19

$t h a n k y o u$
Commented by mathmax by abdo last updated on 12/Aug/19

$y o u a r e w e l c o m e .$
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